The given problem is to find the value of the series:

$\frac{7}{7 \times 10} + \frac{7}{10 \times 13} + \cdots + \frac{7}{49 \times 52} + \frac{7}{52 \times 55}$

To evaluate this, let's examine each term in the series. Notice that each term has the form:

$\frac{7}{n(n+3)}$

where $n$ follows a pattern of values: $7, 10, 13, \ldots, 52$.

Step 1: Simplify Each Term

We can simplify each term $\frac{7}{n(n+3)}$ by using partial fraction decomposition:

$\frac{7}{n(n+3)} = \frac{A}{n} + \frac{B}{n+3}$

Multiplying through by $n(n+3)$ gives:

$7 = A(n+3) + Bn$

Expanding and solving for $A$ and $B$:

$7 = An + 3A + Bn$

Equating terms, we get:

1. $A + B = 0$
2. $3A = 7$

From $3A = 7$, we find $A = \frac{7}{3}$. Substituting into $A + B = 0$, we get $B = -\frac{7}{3}$.

Thus:

$\frac{7}{n(n+3)} = \frac{7}{3} \left( \frac{1}{n} - \frac{1}{n+3} \right)$

Step 2: Rewrite the Series

Using this decomposition, the series becomes:

$\frac{7}{3} \left( \frac{1}{7} - \frac{1}{10} + \frac{1}{10} - \frac{1}{13} + \cdots + \frac{1}{49} - \frac{1}{52} + \frac{1}{52} - \frac{1}{55} \right)$

Step 3: Observe the Telescoping Nature

Notice that the terms cancel out in a telescoping manner, leaving only the first and the last terms:

$\frac{7}{3} \left( \frac{1}{7} - \frac{1}{55} \right)$

Step 4: Simplify the Remaining Expression

Calculating the remaining expression:

$\frac{7}{3} \left( \frac{1}{7} - \frac{1}{55} \right) = \frac{7}{3} \left( \frac{55 - 7}{7 \times 55} \right) = \frac{7}{3} \cdot \frac{48}{385}$

Simplifying further:

$= \frac{7 \times 48}{3 \times 385} = \frac{336}{1155} = \frac{112}{385}$

Final Answer

The value of the series is:

$\frac{112}{385}$

Would you like a further breakdown of these steps?

Here are some related questions for further exploration:

1. How would you approach a similar series with different constants in the denominator?
2. Can you generalize the formula for the sum of a series in the form $\frac{k}{n(n+m)}$?
3. How would the answer change if the series extended to infinity?
4. What would happen if the numerator had a different value than 7?
5. Can you derive a telescoping series for other patterns of terms?

Tip: Telescoping series often simplify complex sums by allowing terms to cancel out, leaving only a few terms for calculation.